1981 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

If

\sqrt{x+2}=2,

then

(x+2)^2

equals

\sqrt2

2

4

8

16

Concepts:radical substitution

Difficulty rating: 900

Small hint:

The equation already gives the value of

x+2

Big hint:

Square the known value once more

Solution:

From

\sqrt{x+2}=2,

we get

x+2=4.

Therefore

(x+2)^2=4^2=16.

Therefore, the correct answer is E.

2.

Point

E

is on side

AB

of square

ABCD.

If

EB

has length one and

EC

has length two, then the area of the square is

\sqrt3

\sqrt5

3

2\sqrt3

5

Concepts:square (geometry)Pythagorean Theorem area

Difficulty rating: 1160

Small hint:

Triangle

EBC

is a right triangle

Big hint:

The unknown leg

BC

is also the side length of the square

Solution:

Let the square’s side length be

s.

In right triangle

EBC,

s^2+1^2=2^2.

Hence

s^2=3,

which is exactly the area of the square.

Therefore, the correct answer is C.

3.

For

x\ne0,

\frac1x+\frac1{2x}+\frac1{3x}

equals

\frac1{2x}

\frac1{6x}

\frac5{6x}

\frac{11}{6x}

\frac1{6x^3}

Concepts:fraction algebraic manipulation

Difficulty rating: 920

Small hint:

Use

6x

as a common denominator

Big hint:

Add the resulting numerators

6,

3,

and

2

Solution:

Using the common denominator

6x

gives

\begin{aligned} \frac1x+\frac1{2x}+\frac1{3x} &=\frac6{6x}+\frac3{6x}\\ &\quad+\frac2{6x}\\ &=\frac{11}{6x}. \end{aligned}

Therefore, the correct answer is D.

4.

If three times the larger of two numbers is four times the smaller and the difference between the numbers is

8,

then the larger of the two numbers is

16

24

32

44

52

Concepts:linear equation ratio and proportion

Difficulty rating: 980

Small hint:

Let

L

and

S

denote the larger and smaller numbers

Big hint:

Use

3L=4S

together with

L-S=8

Solution:

The relation

3L=4S

gives

S=\frac{3L}{4}.

Thus

L-S=L-\frac{3L}{4}=\frac L4=8,

so

L=32.

Therefore, the correct answer is C.

5.

In trapezoid

ABCD,

sides

AB

and

CD

are parallel, and diagonal

BD

and side

AD

have equal length. If

\angle DCB=110^\circ

and

\angle CBD=30^\circ,

then

\angle ADB=

80^\circ

90^\circ

100^\circ

110^\circ

120^\circ

Concepts:trapezoid parallel lines isosceles triangle angle chasing

Difficulty rating: 1360

Small hint:

Use the parallel bases to find the full interior angle at

B

Big hint:

Then use

AD=BD

in triangle

ABD

Solution:

Since

AB\parallel CD,

consecutive interior angles give

\angle ABC=180^\circ-110^\circ=70^\circ.

Hence

\angle ABD=70^\circ-30^\circ=40^\circ.

Because

AD=BD,

triangle

ABD

has

\angle DAB=\angle ABD=40^\circ.

Therefore

\begin{aligned} \angle ADB &=180^\circ-40^\circ-40^\circ\\ &=100^\circ. \end{aligned}

Therefore, the correct answer is C.

6.

If

\frac{x}{x-1}=\frac{y^2+2y-1}{y^2+2y-2},

then

x

equals

y^2+2y-1

y^2+2y-2

y^2+2y+2

y^2+2y+1

-y^2-2y+1

Concepts:rational equation algebraic manipulation

Difficulty rating: 1210

Small hint:

Notice that each denominator is one less than its numerator

Big hint:

Set

A=y^2+2y-1

and compare

\frac{x}{x-1}

with

\frac A{A-1}

Solution:

Let

A=y^2+2y-1.

The equation becomes

\frac{x}{x-1}=\frac{A}{A-1}.

Cross-multiplication gives

xA-x=Ax-A,

so

x=A=y^2+2y-1.

Therefore, the correct answer is A.

7.

How many of the first one hundred positive integers are divisible by all of the numbers

2,

3,

4,

5?

0

1

2

3

4

Concepts:least common multiple divisibility

Difficulty rating: 980

Small hint:

Find the least common multiple of

2,

3,

4,

and

5

Big hint:

Count the multiples of that least common multiple through

100

Solution:

The least common multiple is

\operatorname{lcm}(2,3,4,5)=60.

Among the positive integers through

100,

only

60

is a multiple of

60.

Thus the count is

1.

Therefore, the correct answer is B.

8.

For all positive numbers

x,

y,

z,

the product

\begin{aligned} &(x+y+z)^{-1}\\ &\quad\cdot(x^{-1}+y^{-1}+z^{-1})\\ &\quad\cdot(xy+yz+zx)^{-1}\\ &\quad\cdot\big((xy)^{-1}+(yz)^{-1}+(zx)^{-1}\big) \end{aligned}

equals

x^{-2}y^{-2}z^{-2}

x^{-2}+y^{-2}+z^{-2}

(x+y+z)^{-1}

\frac1{xyz}

\frac1{xy+yz+zx}

Concepts:algebraic manipulation fraction

Difficulty rating: 1630

Small hint:

Rewrite each sum of reciprocals over a common denominator

Big hint:

The two reciprocal sums introduce factors

xy+yz+zx

and

x+y+z

Solution:

We have

x^{-1}+y^{-1}+z^{-1}=\frac{xy+yz+zx}{xyz}

and

\begin{aligned} &(xy)^{-1}+(yz)^{-1}\\ &\quad+(zx)^{-1} =\frac{x+y+z}{xyz}. \end{aligned}

All symmetric sum factors cancel, leaving

\frac{1}{x^2y^2z^2}=x^{-2}y^{-2}z^{-2}.

Therefore, the correct answer is A.

9.

In the adjoining figure,

PQ

is a diagonal of the cube. If

PQ

has length

a,

then the surface area of the cube is

2a^2

2\sqrt2a^2

2\sqrt3a^2

3\sqrt3a^2

6a^2

Concepts:cube geometry diagonal surface area

Difficulty rating: 1280

Small hint:

Relate the cube’s space diagonal to its side length

Big hint:

Substitute the side length into the formula for six square faces

Solution:

If the side length is

s,

then the space diagonal is

s\sqrt3=a,

so

s^2=\frac{a^2}{3}.

The surface area is

6s^2=6(\frac{a^2}{3})=2a^2.

Therefore, the correct answer is A.

10.

The lines

L

and

K

are symmetric to each other with respect to the line

y=x.

If the equation of line

L

is

y=ax+b

with

a\ne0

and

b\ne0,

then the equation of

K

is

y=

\frac1a x+b

-\frac1a x+b

-\frac1a x-\frac ba

\frac1a x+\frac ba

\frac1a x-\frac ba

Concepts:coordinate geometry reflection (geometry)linear equation

Difficulty rating: 1300

Small hint:

Reflection across

y=x

interchanges the two coordinates

Big hint:

Swap

x

and

y

in the original equation, then solve for

y

Solution:

Interchanging

x

and

y

in

y=ax+b

gives

x=ay+b.

Solving,

y=\frac{x-b}{a}=\frac1a x-\frac ba.

Therefore, the correct answer is E.

11.

The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the sides could have length

22

58

81

91

361

Concepts:Pythagorean Triple arithmetic sequence divisibility

Difficulty rating: 1360

Small hint:

Represent the three equally spaced sides as

s-d,

s,

s+d

Big hint:

Every such integral right triangle is a multiple of a

3

-

4

-

5

triangle

Solution:

If the sides are

s-d,

s,

s+d,

the Pythagorean equation gives

(s-d)^2+s^2=(s+d)^2,

hence

s=4k

and

d=k

for some positive integer

k.

Thus the sides are

3k,

4k,

5k.

Of the choices,

81=3\cdot27

can occur.

Therefore, the correct answer is C.

12.

If

p,

q,

and

M

are positive numbers and

q\lt100,

then the number obtained by increasing

M

by

p\%

and decreasing the result by

q\%

exceeds

M

if and only if

p\gt q

p\gt\frac q{100-q}

p\gt\frac q{1-q}

p\gt\frac{100q}{100+q}

p\gt\frac{100q}{100-q}

Concepts:percentage inequality algebraic manipulation

Difficulty rating: 1450

Small hint:

Write the two successive percentage changes as multiplication factors

Big hint:

Use

q\lt100

when dividing by

100-q

Solution:

The final amount is

M\left(1+\frac p{100}\right) \left(1-\frac q{100}\right).

It exceeds

M

exactly when

(100+p)(100-q)\gt10000.

Expanding gives

p(100-q)\gt100q.

Since

100-q\gt0,

this is

p\gt\frac{100q}{100-q}.

Therefore, the correct answer is E.

13.

Suppose that at the end of any year, a unit of money has lost

10\%

of the value it had at the beginning of that year. Find the smallest integer

n

such that after

n

years the unit of money will have lost at least

90\%

of its value. (To the nearest thousandth

\log_{10}3

is

0.477.

)

14

16

18

20

22

Concepts:exponent logarithm inequality

Difficulty rating: 1660

Small hint:

After

n

years, the fraction of value remaining is

0.9^n

Big hint:

Use

\log_{10}0.9=2\log_{10}3-1

Solution:

At least

90\%

lost means

0.9^n\le0.1.

Now

\begin{aligned} \log_{10}0.9 &=\log_{10}9-1\\ &=2(0.477)-1\\ &=-0.046. \end{aligned}

Thus

n(-0.046)\le-1,

so

n\ge21.739\ldots.

The least integer is

22.

Therefore, the correct answer is E.

14.

In a geometric sequence of real numbers, the sum of the first two terms is

7,

and the sum of the first six terms is

91.

The sum of the first four terms is

28

32

35

49

84

Concepts:geometric sequence quadratic

Difficulty rating: 1730

Small hint:

Factor the first-six-term sum into the first-two-term sum times

1+r^2+r^4

Big hint:

Set

t=r^2

and solve

1+t+t^2=13

Solution:

If the first term is

a

and ratio is

r,

then

91=a(1+r)(1+r^2+r^4).

Since

a(1+r)=7,

we have

1+r^2+r^4=13.

With

t=r^2\ge0,

t^2+t-12=0,

so

t=3.

The first four terms sum to

a(1+r)(1+r^2)=7(4)=28.

Therefore, the correct answer is A.

15.

If

b\gt1,

x\gt0,

and

(2x)^{\log_b2}-(3x)^{\log_b3}=0,

then

x

is

\frac1{216}

\frac16

1

6

not uniquely determined

Concepts:logarithm exponent substitution

Difficulty rating: 1860

Small hint:

Take logarithms to base

b

of the two equal positive powers

Big hint:

Let

A=\log_b2,

B=\log_b3,

and

u=\log_bx

Solution:

Let

A=\log_b2,

B=\log_b3,

and

u=\log_bx.

Taking base-

b

logarithms gives

A(A+u)=B(B+u).

Since

A\ne B,

A+B+u=0.

Therefore

u=-\log_b6=\log_b(\frac{1}{6}),

so

x=\frac{1}{6}.

Therefore, the correct answer is B.

16.

The base three representation of

x

is

12112211122211112222.

The first digit (on the left) of the base nine representation of

x

is

1

2

3

4

5

Concepts:number base digits

Difficulty rating: 1450

Small hint:

Since

9=3^2,

group the base-three digits in pairs from the right

Big hint:

Convert the leftmost pair

12_3

into a base-nine digit

Solution:

The representation has an even number of base-three digits, so its leftmost base-nine digit comes from

12_3.

That pair has value

1\cdot3+2=5,

so the first base-nine digit is

5.

Therefore, the correct answer is E.

17.

The function

f

is not defined for

x=0,

but, for all nonzero real numbers

x,

f(x)+2f\left(\frac1x\right)=3x.

The equation

f(x)=f(-x)

is satisfied by

exactly one real number

exactly two real numbers

no real numbers

infinitely many, but not all, nonzero real numbers

all nonzero real numbers

Concepts:functional equation system of equations symmetry (algebra)

Difficulty rating: 1940

Small hint:

Apply the given relation again after replacing

x

by

\frac{1}{x}

Big hint:

Solve the two equations for

f(x)

, then compare its values at

x

and

-x

Solution:

Replacing

x

by

\frac{1}{x}

gives

f(\frac{1}{x})+2f(x)=\frac{3}{x}.

Solving the two linear equations yields

f(x)=\frac2x-x.

Hence

f(-x)=-f(x),

so equality requires

f(x)=0.

Thus

\frac{2}{x}-x=0,

or

x^2=2.

Both

x=\sqrt2

and

x=-\sqrt2

work, giving exactly two real numbers.

Therefore, the correct answer is B.

18.

The number of real solutions to the equation

\frac{x}{100}=\sin x

is

61

62

63

64

65

Concepts:trigonometry counting intersections symmetry

Difficulty rating: 2130

Small hint:

Use odd symmetry and first count the positive solutions

Big hint:

Only positive sine humps below

x=100

can meet the line, with special care for the first hump

Solution:

The equation is odd-symmetric and has the solution

x=0.

On

(0,\pi)

there is one further positive solution. For each

k=1,2,\ldots,15,

the positive sine hump on

(2k\pi,(2k+1)\pi)

rises above

\frac{x}{100}

and then falls below it, giving two solutions. There are no positive solutions beyond

31\pi

, because the next positive hump begins above

100

while

|\sin x|\le1.

Thus there are

1+2(15)=31

positive solutions, the same number negative, and zero itself:

31+31+1=63.

Therefore, the correct answer is C.

19.

In

\triangle ABC,

M

is the midpoint of side

BC,

AN

bisects

\angle BAC,

BN\perp AN,

and

\theta

is the measure of

\angle BAC.

If sides

AB

and

AC

have lengths

14

and

19,

respectively, then length

MN

equals

2

\frac52

\frac52-\sin\theta

\frac52-\frac12\sin\theta

\frac52-\frac12\sin\left(\frac\theta2\right)

Concepts:angle bisector reflection (geometry)midpoint

Difficulty rating: 1860

Small hint:

Extend

BN

through

N

to meet

AC

Big hint:

Use congruence across the angle bisector, then apply the midpoint theorem

Solution:

Extend

BN

to meet

AC

at

E.

The right triangles

ABN

and

AEN

are congruent because

AN

is common and bisects the angle at

A.

Hence

BE

is bisected by

N

and

AE=AB=14.

Since

AC=19,

EC=5.

In triangle

BEC,

points

N

and

M

are the midpoints of

BE

and

BC,

so

MN=\frac{EC}{2}=\frac{5}{2}.

Therefore, the correct answer is B.

20.

A ray of light originates from point

A

and travels in a plane, being reflected

n

times between lines

AD

and

CD,

before striking a point

B

(which may be on

AD

or

CD

) perpendicularly and retracing its path to

A.

(At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for

n=3.

) If

\angle CDA=8^\circ,

what is the largest value

n

can have?

6

10

38

98

There is no largest value.

Concepts:reflection (geometry)angle chasing inequality

Difficulty rating: 1950

Small hint:

Track the acute angle between the ray and each successive reflecting line

Big hint:

Each reflection advances that angle by the wedge angle

8^\circ

Solution:

Let

\theta\gt0

be the initial acute angle between the ray and

AD.

Successive exterior-angle relations increase the corresponding acute angle by

8^\circ

at each reflection. At the final perpendicular strike,

90^\circ=\theta+(8n+8)^\circ.

Thus

\theta=82^\circ-8n^\circ\gt0,

so

n\lt\frac{82}{8}=10.25.

The greatest integer possible is

10,

attained when

\theta=2^\circ.

Therefore, the correct answer is B.

21.

In a triangle with sides of lengths

a,

b,

and

c,

(a+b+c)(a+b-c)=3ab.

The measure of the angle opposite the side of length

c

is

15^\circ

30^\circ

45^\circ

60^\circ

150^\circ

Concepts:law of cosines difference of squares

Difficulty rating: 1450

Small hint:

Expand the left side as a difference of squares

Big hint:

Compare the resulting expression for

c^2

with the law of cosines

Solution:

The condition gives

(a+b)^2-c^2=3ab,

so

c^2=a^2+b^2-ab.

If

\theta

is opposite

c,

the law of cosines says

c^2=a^2+b^2-2ab\cos\theta.

Therefore

2ab\cos\theta=ab,

so

\cos\theta=\frac{1}{2}

and

\theta=60^\circ.

Therefore, the correct answer is D.

22.

How many lines in a three-dimensional rectangular coordinate system pass through four distinct points of the form

(i,j,k),

where

i,

j,

and

k

are positive integers not exceeding four?

60

64

72

76

100

Concepts:3D geometry lattice point basic counting

Difficulty rating: 2040

Small hint:

Four collinear lattice points in this

4

-by-

4

-by-

4

grid must advance by coordinate steps

0

or

\pm1

Big hint:

Count axis-parallel lines, face-diagonal lines, and space diagonals separately

Solution:

There are

3\cdot4^2=48

axis-parallel lines. For face-diagonal directions, choose the pair of varying coordinates in

3

ways, choose one of

2

diagonal slopes, and fix the remaining coordinate in

4

ways, giving

3\cdot2\cdot4=24.

Finally, the cube has

4

space diagonals. Thus the total is

48+24+4=76.

Therefore, the correct answer is D.

23.

Equilateral

\triangle ABC

is inscribed in a circle. A second circle is tangent internally to the circumcircle at

T

and tangent to sides

AB

and

AC

at points

P

and

Q.

If side

BC

has length

12,

then segment

PQ

has length

6

6\sqrt3

8

8\sqrt3

9

Concepts:equilateral triangle tangent circles similarity

Difficulty rating: 1910

Small hint:

Use symmetry to place both circle centers and

T

on the altitude from

A

Big hint:

Relate the smaller circle’s radius to the circumradius using the

30^\circ

half-angle at

A

Solution:

The circumradius of the equilateral triangle is

R=\frac{12}{\sqrt3}=4\sqrt3.

If the smaller radius is

r,

its center lies on the altitude and is

2r

from

A,

because its distance to either side is

r

and the half-angle is

30^\circ.

Internal tangency at the bottom gives

2r=2R-r,

so

r=\frac{2R}{3}=\frac{8\sqrt3}{3}.

The tangency points on the two sides are separated by

PQ=r\sqrt3=8.

Therefore, the correct answer is C.

24.

If

\theta

is a constant such that

0\lt\theta\lt\pi

and

x+\frac1x=2\cos\theta,

then for each positive integer

n,

x^n+\frac1{x^n}

equals

2\cos\theta

2^n\cos\theta

2\cos^n\theta

2\cos n\theta

2^n\cos^n\theta

Concepts:complex number De Moivre’s Theorem trigonometric identity

Difficulty rating: 1940

Small hint:

Rewrite the condition as a quadratic equation in

x

Big hint:

Its two roots are

\cos\theta\pm i\sin\theta

Solution:

The equation is

x^2-2x\cos\theta+1=0,

whose roots are

e^{i\theta}

and

e^{-i\theta}.

Thus

x=e^{\pm i\theta}

and

x^{-1}=e^{\mp i\theta}.

By De Moivre’s theorem,

\begin{aligned} x^n+x^{-n} &=e^{in\theta}+e^{-in\theta}\\ &=2\cos n\theta. \end{aligned}

Therefore, the correct answer is D.

25.

In triangle

ABC

in the adjoining figure,

AD

and

AE

trisect

\angle BAC.

The lengths of

BD,

DE,

and

EC

are

2,

3,

and

6,

respectively. The length of the shortest side of

\triangle ABC

is

2\sqrt{10}

11

6\sqrt6

6

not uniquely determined by the given information

Concepts:angle bisector theorem law of cosines angle chasing

Difficulty rating: 2210

Small hint:

Let the three equal angles be

\alpha

, and apply the angle-bisector theorem to a suitable subtriangle

Big hint:

Equate law-of-cosines expressions for

\cos\alpha

in the three adjacent triangles

Solution:

Let

AB=c,

AD=y,

AC=b,

and

AE=z.

Since

AD

bisects

\angle BAE,

and

AE

bisects

\angle DAC,

the angle-bisector theorem gives

\frac{c}{z}=\frac{2}{3},\qquad \frac{y}{b}=\frac{3}{6},

so

z=\frac{3c}{2}

and

b=2y.

Applying the law of cosines to triangles

ADB,

ADE,

and

AEC

and equating their common

\cos\alpha

values gives

\begin{gathered} 3c^2-2y^2=12,\\ 9c^2-8y^2=-72. \end{gathered}

Hence

y^2=54

and

c^2=40.

The sides are

AB=2\sqrt{10},

BC=11,

and

AC=2\sqrt{54}=6\sqrt6,

so the shortest is

2\sqrt{10}.

Therefore, the correct answer is A.

26.

Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is

\frac16,

independent of the outcome of any other toss.)

\frac13

\frac29

\frac5{18}

\frac{25}{91}

\frac{36}{91}

Concepts:dice (probability)geometric distribution independent events

Difficulty rating: 1680

Small hint:

For Carol to win in a given round, Alice and Bob must fail before Carol succeeds

Big hint:

If all three fail, the process restarts with the same probabilities

Solution:

Carol wins in the first round with probability

(\frac{5}{6})^2(\frac{1}{6})=\frac{25}{216}.

A complete round with no six has probability

(\frac{5}{6})^3=\frac{125}{216}.

Therefore the desired geometric series is

\frac{\frac{25}{216}}{1-\frac{125}{216}} =\frac{25}{91}.

Therefore, the correct answer is D.

27.

In the adjoining figure triangle

ABC

is inscribed in a circle. Point

D

lies on

\overset{\frown}{AC}

with

\overset{\frown}{DC}=30^\circ,

and point

G

lies on

\overset{\frown}{BA}

with

\overset{\frown}{BG}\gt\overset{\frown}{GA}.

Side

AB

and side

AC

each have length equal to the length of chord

DG,

and

\angle CAB=30^\circ.

Chord

DG

intersects sides

AC

and

AB

at

E

and

F,

respectively. The ratio of the area of

\triangle AFE

to the area of

\triangle ABC

is

\frac{2-\sqrt3}{3}

\frac{2\sqrt3-3}{3}

7\sqrt3-12

3\sqrt3-5

\frac{9-5\sqrt3}{3}

Concepts:arc chord area ratio

Difficulty rating: 2260

Small hint:

Use equal chords to identify the relevant equal arcs and isosceles triangles

Big hint:

Normalize

AB=AC=DG=1

and find

AE

from a

30^\circ

-

60^\circ

-

90^\circ

triangle

Solution:

Scale so

AB=AC=DG=1.

The equal-chord arc relations show that triangle

DEC

is

30^\circ

-

60^\circ

-

90^\circ

and

AE=DE.

Put

AE=DE=x.

Then

CE=1-x=\frac{2x}{\sqrt3},

so

x=2\sqrt3-3.

Equal chords also give

AF=FG=EF=\frac{1-x}{2}.

Hence

\begin{aligned} \frac{[AFE]}{[ABC]} &=\frac{x(1-x)}2\\ &=7\sqrt3-12. \end{aligned}

Therefore, the correct answer is C.

28.

Consider the set of all equations

x^3+a_2x^2+a_1x+a_0=0,

where

a_2,

a_1,

a_0

are real constants and

|a_i|\le2

for

i=0,

1,

2.

Let

r

be the largest positive real number which satisfies at least one of these equations. Then

1\le r\lt\frac32

\frac32\le r\lt2

2\le r\lt\frac52

\frac52\le r\lt3

3\le r\lt\frac72

Concepts:polynomial inequality bounding to limit cases

Difficulty rating: 2210

Small hint:

For positive

x,

the coefficients that push a root farthest right take their lower bounds

Big hint:

Bracket the largest root of

x^3-2x^2-2x-2

at two consecutive choice endpoints

Solution:

Write an allowed polynomial as

g.

For

x\ge0,

\begin{aligned} g(x)&=x^3+a_2x^2\\ &\quad+a_1x+a_0\\ &\ge f(x), \end{aligned}

where

f(x)=x^3-2x^2-2x-2.

Thus no positive root can exceed the largest root

\rho

of

f.

Taking

a_2=a_1=a_0=-2

gives

g=f,

so

\rho

itself is attained. Since

\begin{gathered} f\left(\frac52\right)=-\frac{31}{8}\lt0,\\ f(3)=1\gt0, \end{gathered}

we have

\frac{5}{2}\lt\rho\lt3.

Therefore, the correct answer is D.

29.

If

a\gt1,

then the sum of the real solutions of

\sqrt{a-\sqrt{a+x}}=x

is equal to

\sqrt a-1

\frac{\sqrt a-1}{2}

\sqrt{a-1}

\frac{\sqrt{a-1}}2

\frac{\sqrt{4a-3}-1}{2}

Concepts:radical quadratic casework

Difficulty rating: 2040

Small hint:

The principal square root forces

x\ge0

Big hint:

After one squaring, factor a difference involving

\sqrt{a+x}-x

Solution:

Since

x\ge0,

squaring once gives

a-\sqrt{a+x}=x^2.

Let

y=\sqrt{a+x}.

Then

y=a-x^2

and

a=y^2-x,

so

0=(y+x)(y-x-1).

Since

y+x\gt0,

we have

y=x+1,

or

\sqrt{a+x}=x+1.

Therefore

a+x=x^2+2x+1

and

x^2+x+1-a=0.

The only nonnegative root is

x=\frac{\sqrt{4a-3}-1}{2},

and it satisfies the original equation. It is therefore also the sum of all real solutions.

Therefore, the correct answer is E.

30.

If

a,

b,

c,

d

are the solutions of the equation

x^4-bx-3=0,

then an equation whose solutions are

\begin{gathered} \frac{a+b+c}{d^2},\quad \frac{a+b+d}{c^2},\\ \frac{a+c+d}{b^2},\quad \frac{b+c+d}{a^2} \end{gathered}

is

3x^4+bx+1=0

3x^4-bx+1=0

3x^4+bx^3-1=0

3x^4-bx^3-1=0

none of these

Concepts:polynomial Vieta’s Formulas substitution

Difficulty rating: 2170

Small hint:

Use the missing cubic term to find

a+b+c+d

Big hint:

Each listed expression becomes the negative reciprocal of one original root

Solution:

Vieta’s formulas give

a+b+c+d=0.

Thus, for example,

\frac{a+b+c}{d^2}=-\frac1d,

and similarly the new roots are

-\frac{1}{a},

-\frac{1}{b},

-\frac{1}{c},

-\frac{1}{d}.

Put

x=-\frac{1}{r}

in

r^4-br-3=0.

Multiplying by

x^4

gives

1+bx^3-3x^4=0,

or

3x^4-bx^3-1=0.

Therefore, the correct answer is D.